Respuesta :
Answer:
.18 Joules
Explanation:
start by getting the distance from equilibrium
Fs=kx
Fs/k=x
1.2N/4N/m=x
.3m=x
then use the value for the distance from equilibrium in the equation
PEs=1/2kx^2
PEs=1/2(4N/m)(.3m)^2
PEs=.18 J
The elastic potential is directly proportional to the square of compression. The total elastic potential energy is 0.18 J that is is stored in this compressed spring.
To calculate the total elastic potential first, calculate the distance of compression by Hooke's law,
[tex]F=kx[/tex]
Where,
[tex]F[/tex]- force = 1.2 N
[tex]k[/tex]- Spring constant = 4 N/m
Put the value in the formula and solve for [tex]x,[/tex]
[tex]x = \dfrac {0.2\rm \ N}{4 \rm \ N/m}\\\\x = 0.3 \rm \ m[/tex]
Now, use the total elastic potential,
[tex]U=\dfrac 12kx^2[/tex]
Put the values,
[tex]U=\dfrac 12(4\rm \ N/m)(0.3\rm \ m)^2\\\\U=0.18 J[/tex]
Therefore, the total elastic potential energy is 0.18 J that is is stored in this compressed spring.
To learn more about elastic potential:
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